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The velocity $v$ of a particle moving along a straight line when at a distance $x$ from the origin is given by $a+by^{2}=x^{2}$ where $a$ and $b$ are constants. Then the acceleration is

\[\begin{array}{1 1}(1)\frac{b}{x}&(2)\frac{a}{x}\\(3)\frac{x}{b}&(4)\frac{x}{a}\end {array}\]

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$a+bv^2=x^2$
$b.2v. \large\frac{dv}{dt}$$ =2x. \large\frac{dx}{dt}$
$2b.v. \large\frac{dv}{dt} $$=2x.v$
Acceleration $\large\frac{dv}{dt} =\frac{x}{b}$
Hence 3 is the correct answer.
answered May 16, 2014 by meena.p
 

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